Lengths of the sides of a triangle: sufficient and necessary condition?

For any three positive scales, $a,b,c$, what is the sufficient and necessary condition such that they can form a triangle? Is $a+c>b,a+b>c,b+c>a$ enough? Thanks!

Condition

$a=x+y,\ b=y+z,\ c=z+x\land x,y,z\in \mathbb{R^+}$

Necessity

$a=x+y,\ b=y+z,\ c=z+x$

$a+b>c \implies x+2y+z>z+x \implies y>0$

$b+c>a \implies z>0$

$c+a>b \implies x>0$

Sufficiency

$y>0 \implies x+2y+z>z+x \implies a+b>c$

$z>0 \implies y+2z+x>x+y \implies b+c>a$

$x>0\implies z+2x+y>y+z\implies c+a>b$

Assertion: The system of inequality is both necessary and sufficient, for the existence of a triangle with side lengths = a, b and c.

Proof:

-- Necessity

The shortest path between any two points is the straight line connecting such two points.

-- Sufficiency

Rank the three numbers. Without loss of generality, let a be the number that is no smaller than the other two, i.e. a >= b and a >= c.

At any point O, draw two concentric circles with radius a and b respectively. (OK for them to coincide in the case of a = b.)

Draw a horizontal diameter intersecting the outer circle at Al(eft) and Ar(ight), inner circle at Bl and Br.

Then distance AlBl = a-b and AlBr = a+b. Since a-b < c < a+b, then, by continuity, there must exist a point on the inner circle (denoted as B) such that AlB = c.

Further, points Al, O and B cannot possibly be co-linear (on a straight line) as that would imply that a = b + c.

Al-O-B thus forms the desired triangle.

Q.E.D.